Solutions

This is not a trivial consequence of the mean value theorem because it requires that you define a function

$$g(x)= \begin{cases} \lim_{y\to a^+}f(y)=l_a, & x=a, \\ f(x), & a<x<b, \\ \lim_{y\to b^-}f(y)=l_b, & x=b, \end{cases}$$

and then prove that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ in order to apply the theorem.

Now, $g$ is trivially continuous and differentiable on $(a,b)$ since on this interval $g=f$ and $f$ is differentiable on $(a,b)$. We thus need to prove that $g$ is also continuous at $x=a$ and $x=b$. For $x=a$, let $\varepsilon>0$. Because

$$\lim_{y\to a^+}f(y)=l_a,$$

there exists by definition $\delta>0$ such that

$$\forall y,\ 0<y-a<\delta \implies |f(y)-l_a|<\varepsilon.$$

Since $g(y)=f(y)$ for $a<y<b$ (assuming $\delta<b-a$) then the inequality holds for $g(y)$. Moreover, $g(a)=l_a$. Thus,

$$\forall y,\ 0\le y-a<\delta \implies |g(y)-l_a|<\varepsilon$$

and therefore

$$\lim_{y\to a}g(y)=g(a).$$

It can be proved similarly that

$$\lim_{y\to b}g(y)=g(b).$$

With these conditions satisfied, we can apply the mean value theorem which gives the equality

$$\frac{g(b)-g(a)}{b-a}= \frac{\lim_{y\to b^-}f(y)-\lim_{y\to a^+}f(y)}{b-a}=g'(x)=f'(x)$$

for some $x\in(a,b)$.